JEE Mains · Physics · STD 12 - Magnetism and matter
- JEE Mains 2026
- Medium
- Numerical
A short bar magnet placed with its axis at \(30^{\circ}\) with an external field of 800 Gauss, experiences a torque of \(0.016 N . m\). The work done in moving it from most stable to most unstable position is \(\alpha \times 10^{-3} J\). The value of \(\alpha\) is _________ .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 64
Step-by-step Solution
Detailed explanation
\( \tau=\mu B~sin~\theta\Rightarrow0.016=\mu\times B\times\frac{1}{2} \)
\( \Rightarrow\mu=\frac{0.032}{B} \)
\( W_{ext}=U_{f}-U_{i}=\mu B-(-\mu B)=2\mu B \)
\( =2\times\frac{0.032}{B}\times B \)
\( =0.064~J \)
\( \Rightarrow\mu=\frac{0.032}{B} \)
\( W_{ext}=U_{f}-U_{i}=\mu B-(-\mu B)=2\mu B \)
\( =2\times\frac{0.032}{B}\times B \)
\( =0.064~J \)
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